\newproblem{lay:1_3_29}{
   % Problem identification
	 \begin{large}
	  \hspace{\fill}\newline
     \textbf{Lay, 1.3.29}
	 \end{large}
	 \\
   \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

   % Problem statement
   Let $\mathbf{v}_1$, $\mathbf{v}_2$, ..., $\mathbf{v}_k$ be points in $\mathbb{R}^3$ and suppose that for $j=1,2,...,k$ an object of mass $m_j$ is
	 located at point $\mathbf{v}_j$. Physicists call such objects as \textit{point masses}. The total mass of the system of point masses is
	 \begin{center}
		 $m=m_1+m_2+...+m_k$
	 \end{center}
	The \textit{center of gravity} (or \textit{center of mass}) of the system is:
	 \begin{center}
		 $\overline{\mathbf{v}}=\frac{1}{m}(m_1\mathbf{v}_1+m_2\mathbf{v}_2+...+m_k\mathbf{v}_k)$
	 \end{center}
	Compute the center of mass of the system consisting of the following point masses (see figure):
	\begin{center}
		\includegraphics[scale=0.4]{Tema1/lay_1_3_29.eps}
	\end{center}
}{
   % Solution
	Let us calculate the total mass
	 \begin{center}
		 $m=m_1+m_2+m_3+m_4=4+2+3+5=14 g$
	 \end{center}
	Now, the center of mass
	 \begin{center}
		 $\overline{\mathbf{v}}=\frac{1}{m}(m_1\mathbf{v}_1+m_2\mathbf{v}_2+m_3\mathbf{v}_3+m_4\mathbf{v}_4)=\frac{1}{14}\left(
		   4\begin{pmatrix}2\\-2\\4\end{pmatrix}+2\begin{pmatrix}-4\\2\\3\end{pmatrix}+3\begin{pmatrix}4\\0\\-2\end{pmatrix}+5\begin{pmatrix}1\\-6\\0\end{pmatrix}\right)=
			\begin{pmatrix}\frac{17}{14}\\-\frac{17}{7}\\ \frac{8}{7}\end{pmatrix}$
	 \end{center}
}

\useproblem{lay:1_3_29}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
